Question: $\dfrac{ 2j + 9k }{ 3 } = \dfrac{ 7j - 7l }{ -5 }$ Solve for $j$.
Multiply both sides by the left denominator. $\dfrac{ 2j + 9k }{ {3} } = \dfrac{ 7j - 7l }{ -5 }$ ${3} \cdot \dfrac{ 2j + 9k }{ {3} } = {3} \cdot \dfrac{ 7j - 7l }{ -5 }$ $2j + 9k = {3} \cdot \dfrac { 7j - 7l }{ -5 }$ Multiply both sides by the right denominator. $2j + 9k = 3 \cdot \dfrac{ 7j - 7l }{ -{5} }$ $-{5} \cdot \left( 2j + 9k \right) = -{5} \cdot 3 \cdot \dfrac{ 7j - 7l }{ -{5} }$ $-{5} \cdot \left( 2j + 9k \right) = 3 \cdot \left( 7j - 7l \right)$ Distribute both sides $-{5} \cdot \left( 2j + 9k \right) = {3} \cdot \left( 7j - 7l \right)$ $-{10}j - {45}k = {21}j - {21}l$ Combine $j$ terms on the left. $-{10j} - 45k = {21j} - 21l$ $-{31j} - 45k = -21l$ Move the $k$ term to the right. $-31j - {45k} = -21l$ $-31j = -21l + {45k}$ Isolate $j$ by dividing both sides by its coefficient. $-{31}j = -21l + 45k$ $j = \dfrac{ -21l + 45k }{ -{31} }$ Swap signs so the denominator isn't negative. $j = \dfrac{ {21}l - {45}k }{ {31} }$